Let $|G|=p^{n} $, then for every $d$, that $d|p^{n} $, there are cyclic subgroups of order $d$ for group $G$.
These subgroups be as $G_{0} \subseteq G_{1} \subseteq ...\subseteq G_{n} =G$,where $|G_{i} |=p^{i} $. Every $G_{i} $ is a cyclic $p$-group.
My question:
If $c_{p^{i} } $ is the number of $G_{i} $, then what is $c_{p^{i} } $?
I think you're first statement is a little off. Take $Z/2Z \times Z/2Z$. Four elements, but no cyclic subgroup of order four.
Modified a bit regarding the cyclic-ness, I think Sylow's theorems hold the answers you seek.